Talk:Tidal force
Why is the Weyl Tensor (or its simplification of differentiating Newton's force law equation) necessary?
Similar point: Some physicists, e.g., Richard Feynman in Lectures on Physics, vol 1, attribute the ocean bulge on the 'side' of the earth opposite the moon, to the centrifugal effect due to the rotation of the earth around the earth-moon center of mass. This seems to be a more cogent explanation than some mysterious 'cinching at the waist causes the the other parts to bulge' notion, especially where we are dealing with fluids rather than solids. Alternatively, a 'gravity warps the very fabric of space' argument may make some sense here.
Tidal forces causing the bulge on the other side
I'm surprised no one before Feynman seems to have suggested the obvious centrifugal force from the earth's orbit (one rotation per month) around the center of mass of the moon and the earth. Common geography texts and even encyclopedia britannica are missing this.
This centrifugal force is obviously most significant as it is what balances the gravitation pull from the moon. Without it the earth and moon would fall in on each other. I would think Newton took this into account in his 1686 paper?
It is the DIFFERENCE between this centrifugal force and the the moon's gravitational pull on any material object on the earth's surface that make up the tidal forces "ellipsoid" (similarly for the earth-sun "ellipsoid". From the average values that balance each other, the centrifugal force increases linearly with the distance from, and in a direction away from the earth-moon center of mass. The gravitational pull decreases as one over the square of the distance from the moon.
Centrifugal force not necessary
While convienient to ease explanation, centrifugal force (or rotation) is not necessary for tidal forces to occur. If the moon was falling "directly towards" earth, the tidal forces would also be felt. (Of course, as the two bodies come closer, tidal forces varies).
possible, but would the tidal effect on earh as is, be as strong without the centrifugical force? or am i missing the point here
Formula
(William M. Connolley 09:35, 27 Aug 2004 (UTC)) Having pondered this a bit, I don't think I believe the formula. Certainly the one on the page has a typo: its literally meaningless to have the formula immeadiately followed by "<< r". Also, dr is undefined. Check back to before it was TeX'd: a comma is now missing.
But also, I'm not sure the intended formula is accurate either. I think that if the outside grav field is:
- a = GM/R^2
(a = acceleration; M mass of central body; R distance from) then the "tidal force" between the center and edge of a body radius r at diatance R is:
- delta-a = GM ( 1/R^2 - 1/(R-r)^2)
which is then *approximately* equal to:
- delta-a = GM ( 2/R^3 ) . r
if r << R (modulo a sign convention or two...).
- A clear derivation can be found at fr:Discuter:Sphère de Hill.
- Urhixidur 14:47, 2004 Aug 27 (UTC)
A look at the history shows an interesting migration in the formula over time, but it was never actually right; an extra factor of 2 got inserted early on. I rewrote most of the article, and I hope it's right now. Fpahl 21:24, 6 Oct 2004 (UTC)
I think the article needs to explain better why there is a force in two directions instead of simply adding together in one direction. --ShaunMacPherson 03:19, 17 Oct 2004 (UTC)
Additional effect of rotation?
The article says "For two bodies rotating about their barycenter, the variation in centripetal force required for the rotation adds to the tidal force." I do not believe this. There is no "additional" effect of rotation. The high tides on both sides of the Earth are both perfectly well explained by the tidal effects of the Moon's gravity, and there is no additional effect caused by the fact the the Earth and the Moon are in orbit about their barycenter.
Here's a thought experiment. Consider the positions of the Earth and the Moon at a particular moment in time. If the Earth and the Moon had somehow arrived at those positions and were stationary at the time (but of course free-falling towards each other), then the ocean tides on Earth would be exactly the same as normal. Someone else has made this point above under "Centrifugal force not necessary".
Next, suppose that the Earth and the Moon had somehow arrived at those positions but were passing each other at a high speed, way in excess of their actual orbital speeds about their barycenter. Again, the ocean tides on Earth would be exactly the same as normal.
The fact that in reality, the Earth and the Moon are, in a sense, passing each other at just the right speeds to keep them in roughly circular orbits about their barycenter makes no difference to the ocean tides on Earth.
Where do you get the formula
from? It's not consistent with the earlier one:
for .
Also, you say that r is "the distance from the reference body's center along the axis". But surely, in the term, r has to be the distance from the center of rotation, ie. the barycenter. And where does the term come from? You can't say it comes from the fact that the object is moving in a circle. It's the force that makes the object move in a circle, but its origin must be something else.
So I don't see where you got this formula from, and I think this entire section on "Additional effect of rotation" is unnecessary and, moreover, incorrect.
I'm sorry if all this seems to break the rule on civility, but I think this section is wrong - I don't know how else to say it. I've presented my reasoning, let's discuss it.